2 Answers
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The subgroup $\langle x,C_F(x)\rangle$ is a subgroup of a free group, hence is free. The only free groups with nontrivial center are the free groups of ranks $0$ or $1$, and $x\neq 1$ is clearly central in $\langle x, C_F(x)\rangle$. Thus, $\langle x, C_F(x)\rangle$ is cyclic, hence its subgroup $C_f(x)$ is cyclic.

If $F$ is a free group, then there exists a $S \subset F$ such that every element of $F$ can be written uniquely as a product of elements of $S$. Suppose $x \neq 1$. Then $x = (s_1...s_n)^k$ for some $n \geq 1$, $s_i \in S$, and $k$ chosen to be largest. Show that $C_F(x)$ is the cyclic group generated by $s_1...s_n$. Use the fact that everything in $F$ can be written uniquely as product of elements of $S$.

Suppose $F=\mathbb{Z}$. Then wouldn't $S$ be the set of nonnegative prime integers? So consider $12=2^2\times 3$. Then $12$ cannot be written as $(2\times 3)^2$. Am I missing something here?
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math-visitorJun 30 '12 at 9:31

@math-visitor First of all $\mathbb{Z}$ is not even a group under multiplication. You are probably think of $\mathbb{Z}$ with addition which is a free group generated by $1$.
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WilliamJun 30 '12 at 9:38

I think you have to pick $k$ maximal here (i.e. ensure that $s_1\dots s_n$ is not itself a power of anything).
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Ben MillwoodJun 30 '12 at 9:48

Oh, I see. Thanks William! This example is from en.wikipedia.org/wiki/Free_group under free groups. I was wondering how would one know whether one is talking about free groups under multiplication or free groups under addition, and I'm guessing if the group's operation is not explicitly mentioned, one is assuming a group under multiplication...
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math-visitorJun 30 '12 at 9:48

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You can also do it by using the result that a subgroup of a free group is free, but that might be using a sledgehammer to crack a nut!
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Derek HoltJun 30 '12 at 11:07